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<h3 class="heading"><span class="type">Paragraph</span></h3>
<ol class="decimal">
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<p>For the ODE problem</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2\frac{dy}{dt}-y=\sin t, \ \ y(0)=1. \label{ode1}
\end{equation*}
</div>
<ol class="lower-alpha">
<li>
<p>obtain the transformed version as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2[sY(s) -1]-Y(s) = \frac{1}{s^2+1}.
\end{equation*}
</div>
</li>
<li>
<p>Rearrange to get</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
Y(s)=\frac{2s^2+3}{(2s-1)(s^2+1)} =
\frac{A}{2s-1} +\frac{Bs+C}{s^2 +1}
\end{equation*}
</div>
</li>
<li>
<p>Show that <span class="process-math">\(A=\frac{14}{5}, \ B=\frac{-2}{5}, \ C=\frac{-1}{5},\)</span> and take the inverse transform to obtain the final solution to (<code class="code-inline tex2jax_ignore">[cross-reference to target(s) "ode1" missing or not unique]</code>) as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(t)= \frac{7}{5} e^{t/2} -\frac{2}{5} \cos t - \frac{1}{5}\sin t.
\end{equation*}
</div>
</li>
</ol>
</li>
<li>
<p>For the system of ODEs</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
\frac{dy}{dt}-\frac{dx}{dt}+y +2x &amp;=&amp;e^t \label{odesys1} \\
\frac{dy}{dt}+\frac{dx}{dt}-x &amp;=&amp;e^{2t} \\
{\rm Initial \ data:} \ \ \ x(0),y(0)&amp;=&amp;1, \label{odesys}
\end{aligned}
\end{equation*}
</div>
<ol class="lower-alpha">
<li>
<p>transform to obtain $$</p>
<p>$$</p>
</li>
<li>
<p>Rearranging, $$</p>
<p>$$</p>
</li>
<li>
<p>To eliminate <span class="process-math">\(Y(s),\)</span> multiply (<code class="code-inline tex2jax_ignore">[cross-reference to target(s) "s1" missing or not unique]</code>) by <span class="process-math">\(s\)</span> and (<code class="code-inline tex2jax_ignore">[cross-reference to target(s) "s2" missing or not unique]</code>) by <span class="process-math">\((s+1)\)</span> then subtract, and deduce as follows $$</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
Then, by partial fractions,
\end{equation*}
</div>
<p class="continuation">X(s) = +  -  -  .$$</p>
</li>
<li>
<p>From the table of transforms, we can find <span class="process-math">\(x(t)\)</span> as</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
x(t)=e^t + e^{2t} -e^{t/2}\cosh\left(\frac{\sqrt{3}}{2}t\right) -
\frac{1}{\sqrt{3}}e^{t/2}\sinh\left(\frac{\sqrt{3}}{2}t\right).
\end{equation*}
</div>
</li>
<li>
<p>You can find <span class="process-math">\(y(t)\)</span> by differentiating and substituting <span class="process-math">\(\frac{dx}{dt}\)</span> in either of the system equations. Quicker here is to subtract the second equation from the first to obtain</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
-2\frac{dx}{dt}+y+x+2x=e^t-e^{2t}
\end{equation*}
</div>
<p class="continuation">so</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(t)=2\frac{dx}{dt}-3x+e^t-e^{2t}.
\end{equation*}
</div>
</li>
</ol>
</li>
<li>
<p>For the IVP with linear ODEs in variable coefficients,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
ty''-ty'+y=2,\qquad y(0)=2, ~~y'(0)=-4.
\end{equation*}
</div>
<p class="continuation">We have <span class="process-math">\({\mathcal L}[y']=sY(s)-y(0)\text{,}\)</span> then <span class="process-math">\({\mathcal L}[ty']=-\frac{\textrm{d}}{\textrm{d}s}[sY(s)-y(0)]=-[Y(s)+sY'(s)]\text{.}\)</span></p>
<p>Similarly, <span class="process-math">\({\mathcal L}[ty'']=-\frac{\textrm{d}}{\textrm{d}s}[s^2Y(s)-sy(0)-y'(0)]=-[2sY(s)+s^2Y'(s)-y(0)]\text{.}\)</span></p>
<p>Thus, rewrite the differential equation in s-domain:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
-[2sY(s)+s^2Y'(s)-y(0)]+[Y(s)+sY'(s)]+Y(s) &amp;=&amp; \frac{2}{s}\\
(-s^2+s)Y'(s)+2(1-s)Y(s) &amp;=&amp; \frac{2-2s}{s}\\
Y'(s)+\frac{2}{x}Y(s) &amp;=&amp; \frac{2}{s^2}.\\
Y(s)  &amp;=&amp; \frac{2}{s}+\frac{c}{s^2},\quad c\in\mathbb{R}
\end{aligned}
\end{equation*}
</div>
<p class="continuation">The inverse transform gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(t)=2+ct.
\end{equation*}
</div>
<p class="continuation">Match with the second initial condition which we haven’t used yet, the solution to this IVP is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y(t) = 2-4t.
\end{equation*}
</div>
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</ol>
<span class="incontext"><a href="sec8_4.html#p-461" class="internal">in-context</a></span>
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